Design Article
Gain error affects op amp choices
Soufiane Bendaoud, National Semiconductor
7/14/2006 12:04 PM EDT
As an example, let's consider the configuration of Figure 1:

Vout/Vin = 1 + R2/R1. This is true if we consider the op amp to be ideal. But there's more to that than just this simple expression when considering nonideal op amps. The gain is expressed in the form of Equation 1:

β, in this case, is R1/(R1+R2), and where A(s) is the open-loop gain and β is the feedback factor, also defined as the portion of the output being fed back to the input.
Dividing the numerator and denominator of Equation 1 by A(s) gives the following expression: 1/(1/A(s) + β).
We now multiply this expression by 1/β and obtain Equation 2:

To simplify Equation 2 above, we write T = A(s) β.
We then add and subtract the same term T to the numerator of Equation 2.
Rearranging the equation yields Equation 3:

When A(s) approaches infinity, the gain is simply equal to the ideal gain 1/β. When A(s) is less than infinity, however, we must consider the error term. Table 1 depicts an example of various op amps configured in a gain of 100 (β = 1/100) under the same conditions, but with different open-loop gains.

Clearly, the error term is proportional to noise gain (closed-loop gain), but inversely proportional to open-loop gain. Thus it's important to choose a high open-loop gain amp for applications requiring high closed-loop gains.
About the author
Soufiane Bendaoud is a product marketing manager at National Semiconductor Corp. National Semiconductor Corp., Sunnyvale, CA.




Comments
manu-s
10/20/2009 2:54 PM EDT
The error in the first case when AB is 178 should be much less. In the equation 3 we have to take modulus of the denominator - so
ACL = 1/b*(1/(1/sqrt[(AB)^2+1] - because if we don't do that then when AB=1 we get ACL = 1/2B which is not correct because that's the 3db point and at that point we get ACL = 1/(sqrt(2)*B).So when AB=178 then error = 0.01% .
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